HYPERBOLIC SURFACES WITH SUBLINEARLY MANY

SYSTOLES THAT FILL

MAXIME FORTIER BOURQUE

Abstract. For any ε > 0, we construct a closed hyperbolic surface ofgenus g = g(ε) with a set of at most εg systoles that fill, meaning thateach component of the complement of their union is contractible. Thissurface is also a critical point of index at most εg for the systole function,disproving the lower bound of 2g − 1 posited by Schmutz Schaller.

1. Introduction

The moduli space Mg,n of Riemann surfaces of genus g with n punc-tures is an object of great interest to many geometers and topologists. Itencodes all the different complex structures, conformal structures, or hyper-bolic structures (provided 2g + n > 2) supported on a surface with giventopology. The topology of moduli space is largely encoded in its orbifoldfundamental group Γg,n, the mapping class group.

All the torsion-free finite index subgroups of Γg,n have the same cohomo-logical dimension, which is called the virtual cohomological dimension (vcd)of Γg,n. Harer computed the vcd of Γg,n for all g and n ≥ 0 and found aspine (a deformation retract) forMg,n with this smallest possible dimensionwhenever n > 0 [Har86]. When n = 0, the vcd of the mapping class groupis equal to 4g− 5, but a spine of this dimension has yet to be found [BV06,Question 1]. The largest codimension attained so far is equal to 2 [Ji14] (thespace Mg,0 has dimension 6g − 6).

In an unpublished preprint [Thu85], Thurston claimed that the set Xg ofclosed hyperbolic surfaces of genus g ≥ 2 whose systoles fill forms a spineforMg =Mg,0. Recall that a systole is a closed geodesic of minimal length,and a set of curves fills if each component of the complement of their unionis simply connected. Thurston’s proof that Mg deformation retracts ontoXg appears to be difficult to complete [Ji14]. Furthermore, the dimensionof Xg is still not known, mostly because we do not understand which fillingsets of curves can be systoles. Indeed, Thurston writes:

Unfortunately, we do not have a combinatorial characteri-zation of collections of curves which can be the collection ofshortest geodesics on a surface. This seems like a challengingproblem, and until more is understood about how to answerit, there are probably not many applications of the currentresult.

1

2 MAXIME FORTIER BOURQUE

The paper [APP11] provides some partial answers to Thurston’s question.On a closed hyperbolic surface, systoles do not self-intersect and distinctsystoles can intersect at most once. This obvious necessary condition is,however, far from being sufficient. Indeed, a filling set of systoles mustcontain at least ∼ πg/ log g curves [APP11, Theorem 3], but there existfilling collections of ∼ 2

√g geodesics pairwise intersecting at most once

[APP11, Corollary 2]. There is a discrepancy in the opposite direction aswell: a closed hyperbolic surface can have at most Cg2/ log g systoles [Par13,Corollary 1.4], but there exist filling collections with more than g2 geodesicspairwise intersecting at most once [MRT14, Theorem 1.1].

Our main result here is a construction of closed hyperbolic surfaces withfilling sets of systoles containing sublinearly many curves in terms of thegenus. Compare this with [Sch96, SS97] where surfaces with superlinearlymany systoles are found. Though we are still very far1 from the lower boundof πg/ log g, our examples improve upon the previous record of surfaces withfilling sets of 2g systoles [APP11, Section 5] [San18].

Theorem 1.1. For every ε > 0, there exist an integer g ≥ 2 and a closedhyperbolic surface of genus g with a filling set of at most εg systoles.

Near a surface with a filling set of at most εg systoles, Thurston’s set Xgcontains the set of solutions to the same number of equations. This shouldimply that Xg has codimension at most εg in Mg. However, the equationsrequiring the curves to have equal length can be redundant, preventing usfrom applying the implicit function theorem. We only manage to prove thatXg has dimension at least 4g−5 when g is even, but conjecture the following.

Conjecture 1.2. For every ε > 0, there exists an integer g ≥ 2 such thatXg has dimension at least (6− ε)g.

On the other hand, we can prove that a closely related spine, the Morse–Smale complex for the systole function, has dimension much larger than thevirtual cohomological dimension of the mapping class group.

In a series of papers [Sch93, Sch94, SS98, SS99], Schmutz Schaller initiatedthe study of the systole function sys :Mg,n → R+, which records the lengthof any of the shortest closed geodesics on a surface. He proved that thesystole function is a topological Morse function on the Teichmuller spaceTg,n whenever n > 0 [SS99] and Akrout extended this result to n = 0 (andto a more general class of functions) in [Akr03].

Schmutz Schaller constructed a critical point of index 2g−1 for the systolefunction in every genus g ≥ 2 and thought it was “quite possible” that thiswas the smallest achievable index [SS99, p.439]. He verified this hypothesisfor g = 2 by finding all the critical points inM2. If this were true in general,it would imply that the Morse–Smale complex for the systole function hasthe smallest possible dimension 4g − 5 = (6g − 6) − (2g − 1) for a spine ofMg. However, our surfaces show that no such inequality holds.

1The genus g in Theorem 1.1 grows like a tower of exponentials of length roughly 1/ε.

HYPERBOLIC SURFACES WITH SUBLINEARLY MANY SYSTOLES THAT FILL 3

Theorem 1.3. For every ε > 0, there exist an integer g ≥ 2 and a criticalpoint of index at most εg for the systole function on Tg.

Organization. The surfaces arising in Theorems 1.1 and 1.3 are built intwo steps, in a similar fashion as the local maxima from [FBR]. First, in Sec-tion 2, we define a building block (depending on some parameters) which isa surface whose systoles are the boundary components. This surface is mod-elled on a flag-transitive surface map (a generalization of Platonic solids) andcan be cut into isometric right-angled polygons along a collection of geodesicarcs. We then glue building blocks together according to the combinatoricsof certain graphs of large girth with strong transitivity properties in Sec-tion 3. We do this in such a way that the boundaries of the blocks remainsystoles in the larger surface and that the arcs in the blocks connect up toform systoles as well (see Section 4). In Section 5, we show that X/ Isom(X)is isometric to a triangle or a quadrilateral. This easily implies that X is acritical point of the systole function, which we prove in Section 6. Finally,we discuss our failed attempt to prove Conjecture 1.2 in Section 7.

Acknowledgements. I thank the anonymous referee for their useful commentsand corrections.

2. Building blocks

Graphs. A graph is a 1-dimensional cell complex, where there can be mul-tiple edges between two vertices and edges from vertices to themselves. Thevalence of a vertex in a graph is the number of half-edges adjacent to it. Ifevery vertex in a graph has the same valence, then this number is called thevalence of the graph.

Flag-transitive maps. A map M is a graph embedded on a surface S suchthat the closure of each complementary component is an embedded closeddisk (called a face of the map). All maps considered in this paper will beorientable, meaning that the surface S is required to be orientable. If allthe faces of a map M have the same number p of edges and all the verticeshave the same valence q, then M is said to have type {p, q}.

A flag in a map is a triple consisting of a vertex v, an edge e containing v,and a face f containing e. A map-automorphism of M is an automorphismof the underlying graph which can be realized by a homeomorphism of thesurface S. A map is flag-transitive2 if its group of map-automorphisms actstransitively on flags.

Any flag-transitive map has type {p, q} for some p ≥ 1 and q ≥ 2. The fivePlatonic solids are the only flag-transitive maps on the sphere with p, q ≥ 3;their types are {3, 3}, {4, 3}, {3, 4}, {5, 3} and {3, 5}. Beach balls assembledfrom q spherical bigons are flag-transitive maps of type {2, q}.

2These maps are usually called regular, but if we stuck to standard terminology, thisword would be used with five different meanings throughout the paper.

4 MAXIME FORTIER BOURQUE

Maps of large girth. A cycle in a graph is a sequence of oriented edges(e1, . . . , ek) such that the endpoint of ei coincides with the starting pointof ei+1 for every i ∈ Zk. Cycles are considered up to cyclic permutation oftheir edges and reversal of orientation. The length of a cycle is the number ofedges that it uses. A cycle is non-trivial if it cannot be homotoped to a pointby deleting backtracks, that is, consecutive edges (modulo k) with oppositeorientations. The girth of a graph is the length of any of its shortest non-trivial cycles. These shortest non-trivial cycles will be called girth cycles. Agraph of girth at most 2 is often called a multigraph, and a graph of girthlarger than 2 is simple.

The girth of a flag-transitive map M of type {p, q} is at most p since thefaces are non-trivial cycles of length p. If M is finite, then one can actuallyunwrap all the cycles shorter than p by taking a suitable finite normal cover,thereby obtaining a finite flag-transitive map N of girth p [Eva79, Theorem11]. That such covers exist follows from Mal’cev’s theorem on the residualfiniteness of finitely generated linear groups [Mal65].

Theorem 2.1 (Evans). For any p, q ≥ 2, there exists a finite flag-transitivemap of type {p, q} and girth p.

See also [Nv01] and [S01] for constructive proofs of this result.

Regular polygons. Let q ≥ 3. Up to isometry, there exists a uniquepolygon P in the hyperbolic plane with 2q sides of the same length L andall interior angles equal to π/2. We will call P the regular right-angled 2q-gon. By connecting the center of P to the midpoint of a side and one of itsvertices, we obtain a triangle with interior angles π/2, π/4 and π/2q and aside of length L/2 from which we obtain the equation

(2.1) cosh(L/2) = cos(π/2q)/ sin(π/4) =√

2 cos(π/2q)

(see [Bus10, p.454]). We color the sides of P red and blue in such a waythat adjacent sides have different colors.

Lemma 2.2. Any arc α between two disjoint sides in the regular right-angled2q-gon P has length at least L, with equality only if α is a side of P .

Proof. Let α have minimal length among such arcs. Then α must be geodesicand orthogonal to ∂P at its endpoints. These endpoints are separated by msides of P in one direction and n sides in the other, where m + n + 2 = 2qand m ≤ n. First suppose that m > 1. Let d be a main diagonal of P whichis linked with α and has one endpoint at an extremity of one of the twosides of P joined by α. Let z be the intersection point between α and d,and let α± be the two components of α \ {z} labelled in such a way that α+

and d have endpoints in a common side of P . If Rd denotes the reflectionabout d, then the arc γ = α− ∪Rd(α+) has the same length as α and joinstwo disjoint sides of P (because m > 1). By minimality, γ must be geodesicand orthogonal to ∂P , which is absurd. This shows that m = 1, in which

HYPERBOLIC SURFACES WITH SUBLINEARLY MANY SYSTOLES THAT FILL 5

case α is a side of P (the orthogonal segment between two geodesics in thehyperbolic plane is unique when it exists). �

One can also prove this using trigonometry (see [APP11, p.91]).

Gluing regular polygons along maps. LetM be an oriented map of type{p, q} where q ≥ 3. Let P be the unique right-angled regular hyperbolic 2q-gon with sides colored red and blue as above. We now define a hyperbolicsurface B modelled on M . For each vertex v ∈ M , take a copy Pv of P .The blue sides of Pv are labelled in counterclockwise order by the edgesadjacent to v in M , which come with a cyclic ordering from the orientation.For each edge e = {u, v} in M , we glue the polygons Pu and Pv along theirsides labelled e by an orientation-reversing isometry. The resulting surfaceis denoted B and will be called a block in the sequel. The polygons Pv ⊂ Bare its tiles.

Topologically, B is the same as the surface S ⊃M with a hole cut out ineach face. Indeed, if we join the center of each polygon Pv to the midpointsof its blue sides, we obtain an embedded copy of M in B. Since eachPv deformation retracts onto the star M ∩ Pv, the surface B deformationretracts onto M . Each boundary component of B is the concatenation of pred sides of polygons Pv coming from the p vertices v around a face of M .In particular, each boundary component of B has length pL, where L is thepositive number implicitly defined by Equation (2.1).

Lemma 2.3. Let M be a map of type {p, q} and girth p, where p ≥ 2 andq ≥ 3. Then the systoles in B are the boundary geodesics, of length pL.

Proof. Let γ be a systole in B. As explained above, the map M embedsin B as the dual graph to the decomposition into the 2q-gons Pv. Letπ : B → M be the nearest point projection. The image π(γ) must be non-trivial in M since γ is non-trivial in B and π is a deformation retraction.It follows that the combinatorial length of π(γ) in M is at least p. In otherwords, γ intersects at least p tiles Pv, joining distinct blue sides each time.By Lemma 2.2, the length of γ ∩ Pv is at least L for any tile Pv that γintersects. The total length of γ is therefore greater than or equal to pL.If equality occurs, then γ must be a concatenation of red arcs, that is, aboundary geodesic. �

Corollary 2.4. Let M be a map of type {p, q} and girth p, where p ≥ 2and q ≥ 3. Then any arc from a boundary component to itself in B whichcannot be homotoped into ∂B has length strictly larger than pL/2.

Proof. Suppose that α is a non-trivial arc of length at most pL/2 from aboundary geodesic b to itself. The arc α followed by the shorter of the twosubarcs of b between its endpoints is a non-trivial closed curve γ of lengthat most pL in B. The closed geodesic homotopic to γ is strictly shorter,contradicting Lemma 2.3. �

6 MAXIME FORTIER BOURQUE

Lemma 2.5. Let M be a map of type {p, q} and girth p, where p ≥ 2 andq ≥ 3. Then any arc α from ∂B to ∂B which cannot be homotoped into ∂Bhas length at least L, with equality only if α is a blue arc.

Proof. Let α be a geodesic arc from ∂B to ∂B. By Lemma 2.3, we mayassume that α joins consecutive sides of any tile Pv it intersects. Since thestarting point of α in on a red side, it has to next intersect a blue side,and then a red. This means that α is homotopic to a blue arc in a unionPu ∪ Pv of two adjacent tiles. This blue arc is shortest among all arcs inPu ∪ Pv joining the same two sides, as it is orthogonal to the boundary atboth endpoints. �

The above results do not require the map M to be finite or flag-transitive,but we will impose these conditions in the next sections.

3. Gluing graphs

In this section, we explain how to glue blocks together along certain graphsof large girth with large automorphism groups in order to get closed hyper-bolic surfaces with many symmetries and few systoles.

Strict polygonal graphs. A strict polygonal graph is a graph G such thatany embedded path of length 2 in G is contained in a unique girth cycle(where cycles are considered up to cyclic reordering and reversal). Thisnotion was introduced by Perkel in his thesis [Per77]. Examples of strictpolygonal graphs include polygons, the tetrahedron, the dodecahedron, andthe cube of any dimension. See [Ser08] for a short survey on the subject.

Archdeacon and Perkel [AP90] found a way to double the girth of a strictpolygonal graph G (or any graph) by taking an appropriate normal covering

space. The girth cycles in this cover G are precisely those that wrap twicearound a girth cycle in G under the covering map. Repeated applicationsof their construction yield strict polygonal graphs of arbitrarily large girthand constant valence (equal to the valence of G).

Seress and Swartz [SS11, Theorem 3.2] proved that any automorphism of

the base graph G lifts to an automorphism of the girth-doubling cover G.They concluded that if G is vertex transitive, edge transitive, arc transitive

or 2-arc transitive, then so is G. We will need an even stronger transitivityproperty, described in the next paragraph.

Isotropic graphs. The star st(v) of a vertex v in a graph is the set ofhalf-edges adjacent to v. A graph G is locally symmetric if for every vertexv ∈ V (G), any bijection of st(v) can be extended to an automorphism ofG that fixes v. We say that a graph is isotropic if it is vertex transitiveand locally symmetric. To spell it out, G is isotropic if every injectionst(u) ↪→ st(v) between stars in G extends to an automorphism of G.

In an isotropic graph, there is a girth cycle passing through any embeddedpath of length 2, but there can be more than one.

HYPERBOLIC SURFACES WITH SUBLINEARLY MANY SYSTOLES THAT FILL 7

Example 3.1. The Petersen graph P (the quotient of the dodecahedron bythe antipodal involution) is an isotropic graph of valence 3 and girth 5 on10 vertices. However, P is not strict polygonal since every embedded pathof length 2 is contained in two distinct girth cycles in P .

Lubotzky [Lub90] constructed infinitely many isotropic Cayley graphs ofany valence d ≥ 3 and any even girth ≥ 6 (the generators are involutions,allowing the valence to be odd). Since we want better control on the girthcycles of our isotropic graphs, we use the girth-doubling construction of

Archdeacon and Perkel instead. The proof that the girth-doubling cover Gof a graph G is isotropic provided that G is isotropic follows immediatelyfrom [SS11, Theorem 3.2], which states that any automorphism of G lifts to

G, and the fact that the covering G → G is normal, so that its deck groupacts transitively on fibers.

The simplest isotropic strict polygonal graph is a pair of vertices joinedby d ≥ 2 edges. Repeated applications of the girth-doubling construction tothis graph Θ yield a sequence of finite, isotropic, strict polygonal graphs ofany valence and arbitrarily large girth.

Theorem 3.2 (Archdeacon–Perkel, Seress–Swartz). For any d ≥ 2 andn ≥ 1, there exists a finite, isotropic, strict polygonal graph G of valence dand girth 2n. In fact, G can be chosen to be a covering space of the bipartitegraph Θ of valence d on 2 vertices, in which case the girth cycles in G projectto powers of girth cycles in Θ under the covering map.

Gluing. We now explain how to glue copies of the block B from Section 2along a finite isotropic strict polygonal graph G to get a closed hyperbolicsurface X with a small set of systoles that fill.

Let q ≥ 3, let n ≥ 1, and write p = 2n. Let M be a finite flag-transitivemap of type {p, q} and girth p whose existence is guaranteed by Theorem 2.1.

Let B be the block obtained by gluing regular right-angled 2q-gons alongthe map M as in Section 2. Let d be the number of boundary componentsof B, which is is equal to the number of faces in M .

Let G be a finite, isotropic, strict polygonal graph of valence d and girthp = 2n covering the bipartite graph Θ on two vertices as in Theorem 3.2,and let π : G → Θ be a covering map. Let σ : V (Θ) → {−1, 1} andχ : E(Θ) → {1, . . . , d} be bijections, where V (Θ) and E(Θ) are the sets ofvertices and edges of Θ respectively. These induce proper colorings σ ◦ πand χ ◦ π of the vertices and edges of G respectively.

For each v ∈ V (G), letBv be a copy of the block B, equipped with its stan-dard orientation if σ(v) = 1 and with the reverse orientation if σ(v) = −1.Let b1, . . . , bd be the boundary components of B and label the boundarycomponents of any copy Bv in the same way so that the isometric identifi-cation Bv ∼= B preserves the indices of boundary components.

Here is how we define the closed hyperbolic surface X given the abovecombinatorial data. For any edge e = {u, v} in G, glue Bu to Bv by the

8 MAXIME FORTIER BOURQUE

identity map along their j-th boundary component, where j = χ◦π(e). Thesurface X is defined as the quotient of tv∈V (G)Bv by these gluings. Sincethe gluing maps are orientation-reversing, X is an oriented surface. It hasempty boundary since the coloring χ◦π takes all values in {1, . . . , d} on theedges containing a given vertex v, so that all the boundary components ofBv are glued. Lastly, X is compact because M and G are finite.

The main reason for using strict polygonal graphs in this construction isso that the blue arcs in the blocks Bv all close up to curves of the samelength in X.

Lemma 3.3. Any blue arc in a block Bv ⊂ X is part of a closed geodesicof length pL in X.

Proof. Any blue arc αv in Bv connects two boundary geodesics bi and bj .The block Bv is glued to two other blocks Bu and Bw via these boundarycomponents, and there are blue arcs αu ⊂ Bu and αw ⊂ Bw correspondingto αv under the isometric identifications Bu ∼= Bv ∼= Bw. The concatenationαu ∪ αv ∪ αw is geodesic since αv is orthogonal to ∂Bv.

By our convention, the arc αu (resp. αw) connects the boundary compo-nents of Bu (resp. Bw) labelled bi and bj . By repeating the above reflectionprocess with αu or αw instead of αv (and so on), we obtain a bi-infinitepath δ = (. . . , u, v, w, . . .) in the graph G whose edges alternate between thecolors i and j. There is also a bi-infinite geodesic

β = · · · ∪ αu ∪ αv ∪ αw ∪ · · ·

in X obtained by concatenating the corresponding blue arcs.Since G is a strict polygonal graph, the path (u, v, w) is contained in

a unique non-trivial cycle γ of length p (the girth of G). Furthermore,Theorem 3.2 stipulates that γ covers a closed cycle of length 2 in Θ underthe covering map π : G → Θ. This cycle of length 2 is necessarily formedby the edges χ−1(i) and χ−1(j) since π respects the coloring of edges. Thismeans that the edges of γ alternate between the colors i and j, and hencethat the path δ wraps around γ periodically in both directions. In otherwords, δ closes up after p steps. Similarly, the geodesic β is closed and itslength is equal to pL since each of its p subarcs has length L. �

Note that we have not used the hypotheses that M is flag-transitive northat G is isotropic yet. This will come up in Section 5 where we determinethe isometry group of X.

4. Systoles

In this section, we determine and count the systoles in the surface Xconstructed above.

Proposition 4.1. Let X be the surface constructed in Section 3. The sys-toles in X are the red curves and the blue curves. These systoles fill X and

HYPERBOLIC SURFACES WITH SUBLINEARLY MANY SYSTOLES THAT FILL 9

there are 4q(q−2)p(g − 1) of them, where q is the valence of the map M , p is

the girth of M and the gluing graph G, and g is the genus of X.

Proof. Let γ be a systole in X. If γ is contained in a single block Bv ⊂ X,then γ is a red curve (of length pL) by Lemma 2.3. Now assume that γ is notcontained in any block. Then the blocks Bv1 , . . . , Bvk (k ≥ 2) that it visitsdefine a closed cycle s = (v1, . . . vk, v1) in the graph G. First suppose thats is trivial in G. Then s contains at least two backtracks, that is, verticesvj in the sequence such that vj−1 = vj+1. If s backtracks at a vertex u ∈ G,this means that a subarc ω of γ enters and leaves the block Bu via the sameboundary component. By Corollary 2.4, ω has length strictly larger thanpL/2. Since there are at least two disjoint subarcs like this, γ is longer thanpL. We conclude that s is non-trivial in G, so that its length is at leastp, the girth of G. But for each vertex u along s, the corresponding subarcof γ in Bv has length at least L by Lemma 2.5. Thus the total length ofγ is at least pL. If equality occurs, then γ is a concatenation of blue arcs.Conversely, any concatenation of blue arcs has length pL by Lemma 3.3.

The complementary components of the set of systoles in X are preciselythe interiors of the tiles from which the blocks are assembled. In particular,the systoles fill.

The number of systoles in X is equal to the total number of red arcs andblue arcs divided by p. This is because the red arcs are joined in groups of pto form systoles, and similarly for the blue arcs. Each such arc α (either redor blue) belongs to exactly two tiles. The rhombus with one vertex in thecenter of each of these two tiles and diagonal α has area π(q − 2)/q by theGauss–Bonnet formula (it has two right angles and two angles π/q). Theserhombi tile X, which has area 4π(g − 1). Therefore, the number of systolesis 4π(g − 1) divided by π(q − 2)/q, divided by p. �

Recall that in the construction of X we could take any q ≥ 3 and p = 2n

for any n ≥ 1. Given any ε > 0, taking n sufficiently large and any q ≥ 3gives a surface with a filling set of at most εg systoles. This proves Theo-rem 1.1. At the other extreme, the largest number of systoles is obtainedwhen q = 3 and p = 2, which gives 6g−6 systoles. By [Sch93, Theorem 2.8],such a surface has too few systoles to be a local maximum of the systolefunction, but we will see later that it is nevertheless a critical point of lowerindex.

Example 4.2. For any g ≥ 2, if we take the map M to be the bipartitegraph of valence g + 1 on two vertices (as a map on the sphere), then theresulting block B has g+ 1 boundary components. Taking the gluing graphG to be equal to M , we obtain a surface X which is the double of B acrossits boundary. The genus of X is then equal to g. Since q = g+ 1 and p = 2,the number of systoles is 2g+ 2 according to the formula in Proposition 4.1.Removing any two intersecting systoles leaves a filling set of 2g systoles.

10 MAXIME FORTIER BOURQUE

This example was previously described in [SS99, Theorem 36] and [APP11,Section 5] and was the starting point of this paper.

Remark 4.3. We could allow the graphs G and M to have different girths pand r by replacing the polygons P in the blocks to be semi-regular with sidelengths Lblue and Lred satisfying pLblue = rLred. A version of Proposition 4.1still holds for this generalization, with the count of systoles coming to

2q

(q − 2)

(1

p+

1

r

)(g − 1).

All one has to do is change Lemma 2.2 to say that the distance between anytwo blue sides is at least Lred and the distance between any two red sides isat least Lblue, and modify the other lemmata accordingly.

5. Isometries

In this section, we determine the isometry group of the surface X upto index 2. Recall that the blocks Bv ⊂ X (where v ∈ V (G)) are tiledby regular right-angled 2q-gons Pu (where u ∈ V (M)). By connecting thecenter of each polygon Pu to the midpoints of its edges with geodesics, weobtain a tiling Q of X by (2, 2, 2, q)-quadrilaterals (i.e., quadrilaterals withthree right angles and one angle equal to π/q). Since any isometry of Xpreserves the set of systoles, it permutes the complementary polygons Puand therefore the quadrilaterals in Q. In fact, any quadrilateral can be sentto any other by an isometry.

Proposition 5.1. The isometry group of X acts transitively on the quadri-laterals in the tiling Q.

Proof. The hypothesis that M is flag-transitive implies that the isometrygroup of B acts transitively on its (2, 2, 2, q)-quadrilaterals. This is becausethere is a one-to-one correspondence between the flags in M and the quadri-laterals in B. The correspondence works as follows. Recall that M naturallyembeds in B, connecting the centers of polygons P to their blue sides. Aflag in M is the same as a half-edge e together with a choice of a face fcontaining e, either on the left or the right. In the tiling of B by quadri-laterals, there are exactly two quadrilaterals that have e as an edge. Theside of e on which f lies determines which quadrilateral to pick. Since anymap-automorphism of M can be realized as an isometry of B and M isflag-transitive, the claim follows.

Let v ∈ V (G) and let φ : Bv → Bv be an isometry. We claim that φextends to an isometry Φ of X. First, the isometry φ induces a permutationτ on {1, . . . , d} such that φ sends the boundary component bi of Bv tothe component bτ(i) for every i. Now the edges adjacent to v in G arecolored with the numbers {1, . . . , d} according to the coloring χ ◦ π. Thusthe permutation τ induces a bijection on the star of v. Since G is locallysymmetric, this bijection can be extended to an automorphism ψ of G. If

HYPERBOLIC SURFACES WITH SUBLINEARLY MANY SYSTOLES THAT FILL 11

x ∈ Bu ⊂ X, then define Φ(x) to be the point φ(x) in Bψ(u), where we usethe canonical identifications Bw ∼= Bv to transport the action of φ onto anyblock. This map is well-defined, for if x ∈ Bu ∩ Bv then x belongs to theboundary component labelled i = χ◦π({u, v}) of Bu and Bv. By definition,φ(x) belongs to the τ(i)-th boundary component of Bv. Now Bψ(u) andBψ(v) are glued along their boundary component labelled χ ◦ π(ψ({u, v})).This number equals τ(i) provided that the automorphism ψ is chosen to be alift of the automorphism of Θ induced by τ , and this is possible according to[SS11, Theorem 3.2]. The map Φ is an isometry since it is a locally isometryas well as a bijection.

Similarly, any automorphism ψ of G which preserves the coloring χ ◦ πdefines an isometry Ψ of X by sending x ∈ Bv to the corresponding x inBψ(v). This simply shuffles the blocks around, acting by the identity map onthe blocks. Note that the group of such automorphisms ψ acts transitivelyon the vertices of G.

Combining these two types of isometries gives the desired result. In orderto send a quadrilateral Q ⊂ Bu to another quadrilateral Q′ ⊂ Bv, first applyan isometry Ψ as in the previous paragraph to send Bu to Bv. Then moveΨ(Q) to Q′ via an isometry Φ of the first type, preserving the block Bv. �

Since there are at most two isometries of X sending one quadrilateral toanother, this determines the isometry group of X up to index 2. We canreformulate this as follows. Subdivide Q further into a tiling T by (2, 4, 2q)-triangles by bisecting the quadrilaterals at their smallest angle. Then theisometry group of X may or may not act transitively on the tiles of Tdepending on the graphs M and G used to construct X.

In Example 4.2, the isometry group of X acts transitively on these trian-gles, but that is not the case in general. That is, there can be an asymmetrybetween the red and blue curves in X. For example, let M be the flag-transitive map of type {4, 4} obtained by subdividing the square torus intoa 5× 5 grid (so that B is a torus with 25 holes) and let G be the 1-skeletonof the 25-dimensional cube. Then each component of X \ {blue curves} is atorus with 16 boundary components corresponding to a 4-dimensional sub-cube of G, while the complementary components of the red curves are theblocks with 25 boundary curves each. In this case, no isometry of X caninterchange the two families of systoles.

6. Critical point and index

A real-valued function f on an n-dimensional manifold M is a topologicalMorse function if for every p ∈ M , there is an open neighborhood U ofp and an injective continuous map φ : U → Rn with φ(p) = 0 such thatf ◦ φ−1 − f(p) takes either the form

(x1, . . . , xn) 7→ x1

12 MAXIME FORTIER BOURQUE

or

(x1, . . . , xn) 7→ −j∑i=1

x2i +n∑

i=j+1

x2i

for some j ∈ {0, . . . , n}. In the first case, p is an ordinary point and in thesecond case p is a critical point of index j. Critical points of index 0 and nare local minima and maxima respectively.

Let g ≥ 2 and let Tg be the Teichmuller space of marked, connected,oriented, closed, hyperbolic surfaces of genus g. This space is a smoothmanifold diffeomorphic to R6g−6. The systole sys(Y ) of a surface Y ∈ Tgis the length of any of its shortest closed geodesics. As mentionned in theintroduction, Akrout [Akr03] proved that sys : Tg → R+ is a topologicalMorse function.

Let Y ∈ Tg and let S be the set of (homotopy classes of) systoles in Y .For each α ∈ S and Z ∈ Tg, we let `α(Z) be the length of the unique closedgeodesic homotopic to α in Z. These functions are differentiable on Tg andwe denote their differentials by d`α.

Definition 6.1. The point Y ∈ Tg is eutactic if for every v ∈ TY Tg, thefollowing implication holds: if d`α(v) ≥ 0 for every α ∈ S, then d`α(v) = 0for every α ∈ S. The rank of a eutactic point Y is the dimension of theimage of the linear map (d`α)α∈S : TY Tg → RS .

With these definitions, we have the following characterization of the cri-tical points of sys [Akr03, Theorem 1].

Theorem 6.2 (Akrout). The critical points of index j of the systole functionare the eutactic points of rank j.

We can now show that the surface X constructed in Section 3 is a criticalpoint of sys and give an upper bound for its index.

Proposition 6.3. Let X be as in Section 3. Then X is a critical point ofindex at most 4q

(q−2)p(g − 1) for the systole function.

Proof. Let S be the set of systoles of X (the red curves and the blue curves).Suppose that v ∈ TXTg is such that d`α(v) ≥ 0 for every α ∈ S and let

w =∑

f∈Isom(X)

f∗v.

Then

(6.1) d`α(w) =∑

f∈Isom(X)

d`α(f∗v) =∑

f∈Isom(X)

d`f(α)(v) ≥ d`α(v) ≥ 0

for every α ∈ S. On the other hand, w is the lift to X of a deformation ofthe quotient orbifold Q = X/ Isom(X). By Proposition 5.1, Q is either a(2, 4, 2q)-triangle or a (2, 2, 2, q)-quadrilateral. If Q is a triangle, then w = 0so that d`α(w) and d`α(v) are both zero by Equation (6.1), for every α ∈ S.

HYPERBOLIC SURFACES WITH SUBLINEARLY MANY SYSTOLES THAT FILL 13

If Q is a quadrilateral, then its deformation space is 1-dimensional. This isbecause any (2, 2, 2, q)-quadrilateral is determined by the lengths a and b ofthe two sides disjoint from the angle π/q, which satisfy the relation

sinh a sinh b = cos(π/q)

(see [Bus10, p.454]). This equation implies that the lengths of the red curvesand the blue curves in S have opposite derivatives in the direction of w. Sincethe derivatives are non-negative, they must all be zero. We conclude thatd`α(v) = 0 for every α ∈ S from Equation (6.1). This shows that X iseutactic. The number of systoles in X is a trivial upper bound for the rankof X, and this number is equal to 4q

(q−2)p(g − 1) by Proposition 4.1. �

Once again, by taking p sufficiently large we obtain critical points of indexat most εg for any ε > 0, thereby proving Theorem 1.3. This disproves thepossibility envisaged by Schmutz Schaller [SS99, p.410] that the minimalindex were 2g − 1.

7. Deformations preserving the systoles

Let Xg be the subset of Tg whose systoles fill. We would like to show thatXg has relatively small codimension in Tg. By Proposition 4.1, the systolesof any surface X constructed in Section 3 fill (recall that X depends onseveral parameters). Let S be the set of systoles in X. If we deform X insuch a way that the curves in S remain of equal length, then these curveswill still be the systoles for sufficiently small deformations. This is becausethe second shortest curve on X is longer by a definite amount and lengthvaries continuously.

In other words, the intersection between the inverse image of the diagonal∆ ⊂ RS by the map (`α)α∈S : Tg → RS and a small neighborhood of X iscontained in Xg. One might be tempted to conclude directly that Xg hascodimension at most |S|−1 in Tg. The subtlety is that the image of (`α)α∈Sis not necessarily transverse to ∆. Indeed, the rank of X can be strictlyless than |S| − 1. For instance, the surface in Example 4.2 has rank 2g − 1according to [SS99, Theorem 36], while |S| = 2g + 2.

To remedy this, one could try to get rid of redundant equations, i.e., tofind a filling subset of curves R ⊂ S for which the differential (dX`α)α∈Ris surjective and apply the implicit function theorem. The problem is thateven if the curves inR stay of equal length, the curves in S\R might becomeshorter and so the systoles might not fill anymore.

Another approach would be to find a nearby surface Xθ which has thesame set of systoles as X, and hope that the differential (dXθ`α)α∈S has fullrank there. Below we will describe a 1-dimensional family of deformationsof X with the same systoles. This fixes the issue of rank in some (but notall) cases. A similar idea was used in [San18] to find a path of surfaces inXg with 2g systoles.

14 MAXIME FORTIER BOURQUE

A 1-dimensional deformation. Recall that X is assembled from right-angled regular 2q-gons P whose sides are colored alternatingly red and blue,where q ≥ 3. Given any θ ∈ (0, π), there exists a unique polygon Pθ with2q equal sides and interior angles alternating between θ and π − θ (startwith a triangle with angles π/q, θ/2 and (π − θ)/2 and reflect repeatedlyacross the two sides at angle π/q). To fix ideas, let us say that θ is thecounter-clockwise (interior) angle from a red side to a blue side when goingclockwise around Pθ and π − θ is the angle from blue to red. Now replaceall the polygons P in X by Pθ while keeping the same gluing combinatorics.By construction, the total angle around vertices of the resulting tiling is 2πso the deformed surface Xθ is still a closed hyperbolic surface. Moreover,the red sides still line up to form closed geodesics and similarly for the bluesides. These closed geodesics all have equal length, namely, p times the sidelength of Pθ. As long as θ is close enough to π/2, these curves will remainthe systoles.

The goal is then to show that the linear map (dXθ`α)α∈S has full rankwhen θ 6= π/2. We can do this for some small examples (see below), but wedo not know how to handle surfaces with complicated gluing graphs of largegirth. We present examples with full rank for girth 2 and 3 below.

Computing the rank. To prove that the derivative of lengths has fullrank, it suffices to find a set of tangent vectors {vβ}β∈S to Teichmullerspace for which the square matrix (dXθ`α(vβ))α,β∈S has non-zero determi-nant. For this, we can choose each vector vβ to be the Fenchel-Nielsen twistdeformation (i.e., left earthquake) around the curve β. The cosine formulaof Wolpert [Wol81] and Kerckhoff [Ker83] then says that

d`α(vβ) =∑p∈α∩β

cos∠p(α, β)

whenever α and β are transverse, where ∠p(α, β) is the counter-clockwiseangle from α to β at the point p. In our case, two distinct curves α, β ∈ Sintersect at most once, with angle θ from red to blue or π − θ from blue tored. If we split the rows and columns of D = (dXθ`α(vβ))α,β∈S by color weget a block matrix of the form

D = cos θ

(0 A−Aᵀ 0

)where A is the matrix of zeros and ones recording which red curves intersectwhich blue curves. If θ 6= π/2, then D has full rank if and only if the matrix

D =

(0 AAᵀ 0

)does. This matrix is the adjacency matrix of some graph IS , namely, thegraph whose vertices are the systoles of X and where two vertices are joinedby an edge if and only if the corresponding systoles intersect.

HYPERBOLIC SURFACES WITH SUBLINEARLY MANY SYSTOLES THAT FILL 15

The determinant of the adjacency matrix of a graph counts somethingcombinatorial on the graph. Indeed, according to [Har62] we have

det(D) =∑J⊂IS

(−1)#{even components of J}2#{cycles in J}

where the sum is over all spanning subgraphs of J ⊂ IS (subgraphs contain-ing all vertices) which are elementary, meaning that their components areeither edges or embedded cycles. The even components are those with aneven number of vertices. In our case, IS is bipartite so that all its cycles areeven.

We are now ready to give some examples where D is invertible.

Examples of girth 2. The first family of examples comes from Exam-ple 4.2. In that example, the red and blue curves form a chain, that is, ISis a cycle of length 2g + 2. It follows that IS has exactly three elementaryspanning subgraphs: IS itself, and two subgraphs obtained by deleting everyother edge in IS . If g = 2m is even, then IS has 4m+ 2 edges and

(7.1) det(D) = (−1)121 + (−1)2m+120 + (−1)2m+120 = −4 6= 0.

Alternatively, one could compute the determinant of D by using the factthat A is a circulant matrix in this case.

The fact that D has non-zero determinant implies that the derivativeof ` = (`α)α∈S : Tg → RS has full rank at Xθ whenever θ 6= π/2. Bythe implicit function theorem, near Xθ we have that `−1(∆) is a smoothsubmanifold of codimension

|S| − 1 = (2g + 2)− 1 = 2g + 1,

hence of dimension 4g − 7. As explained earlier, `−1(∆) intersected with asufficiently small ball around Xθ is contained in Xg. We have thus provedthat Xg has dimension at least 4g−7 when g is even. We can push the proofa little further to obtain the following.

Theorem 7.1. For every even g ≥ 2, the set Xg ⊂ Tg of closed hyperbolicsurfaces of genus g whose systoles fill contains a cell of dimension 4g − 5.

Proof. Let X be the surface of genus g from Example 4.2 and let

S = {α1, . . . , α2g+2}

be its set of systoles labelled in such a way that αj intersects αj−1 andαj+1 for every j, where the indices are taken modulo 2g + 2. Let Xθ bethe deformation of X described above, where θ is close enough to π/2 sothat its sets of systoles is still equal to S. By Equation (7.1), the map` = (`α)α∈S : Tg → RS is a submersion at the point Xθ. In particular, `is open in a neighborhood of Xθ. This implies that there exist surfaces Yarbitrarily close to Xθ such that

`α1(Y ) = · · · = `α2g(Y )

16 MAXIME FORTIER BOURQUE

and such that these lengths are strictly less than `α2g+1(Y ) and `α2g+2(Y ). IfY is close enough to Xθ, then its set of systoles is a subset of S by continuityof the length functions. Therefore, there is a sequence Yn converging to Xθ

such that the systoles in Yn are given by the set R = {α1, . . . , α2g}.If n is large enough, then the square matrix (dYn`α(vβ))α,β∈R will have

non-zero determinant. Indeed, in the limit the matrix has the form

(dXθ`α(vβ))α,β∈R = cos θ

(0 B−Bᵀ 0

)

which is invertible because

(0 BBᵀ 0

)is the adjacency matrix of a tree IR

with an even number of vertices. Up to sign, its determinant is the numberof perfect matchings (spanning subgraphs whose components are edges) inIR, which is equal to one. Furthermore, the entries of (dY `α(vβ))α,β∈Rdepend continuously on the surface Y in the same way that the angles ofintersection between geodesics do.

Let Y = Yn for any such large enough n. Then the systoles in Y aregiven by the set R and the map (`α)α∈R : Tg → RR is a submersion at Y .By the implicit function theorem, the inverse image of the diagonal by thismap is a submanifold of codimension 2g − 1 near Y . Since the curves in Rfill, a small neighborhood of Y in this submanifold is contained in Xg. Thecurves in R fill because the complement of the curves in S is a union of fourpolygons which meet at the intersection of α2g+1 and α2g+2. Adding thesetwo curves fuses the four polygons into a single one. �

When g is odd, the matrix D is singular, but this does not necessarilyimply that the image of (`α)α∈S is not transverse to the diagonal.

An example of girth 3. Next, we present an example of genus g = 6 wherethe underlying graphsM andG for the surfaceX have girth 3 and the matrix

D is non-singular. Let M be the 1-skeleton of a regular tetrahedron (as amap of type {3, 3} on the sphere) and let B be the corresponding block.This is a sphere with 4 holes and tetrahedral symmetry. Although this doesnot fit in the theory of Section 3, it is possible to glue five copies of B alongthe complete graph K5 in such a way that the blue arcs connect up in groupsof three to form closed geodesics. To see this, it is convenient to draw K5

in R3 with a 3-fold symmetry as in Figure 1.The tetrahedral pieces are glued as suggested by the figure, in the simplest

possible way (without twist). By inspection, the blue arcs connect in groupsof three. The proof of Proposition 4.1 applies without change to show thatthe systoles in X are the red curves and the blue curves. The genus of X isequal to the number of edges in the complement of any spanning tree in K5,which is 10 − 4 = 6. Let us label the red curves from 1 to 10 and the bluecurves from a to j as in Figure 1 (the red curves correspond to the edges in

HYPERBOLIC SURFACES WITH SUBLINEARLY MANY SYSTOLES THAT FILL 17

K5 X

1

23

4 5

6

7 8 9

10

ab

c

d

ef

gh

i

j

Figure 1. An embbeding of K5 in R3 and the correspondinggluing of tetrahedral blocks.

K5). Then the intersection matrix A is given by

A =

1 0 0 1 0 1 0 0 0 00 1 0 1 1 0 0 0 0 00 0 1 0 1 1 0 0 0 00 0 0 1 0 0 1 0 0 10 0 0 0 1 0 0 1 0 10 0 0 0 0 1 0 0 1 11 0 0 0 0 0 1 0 1 00 1 0 0 0 0 1 1 0 00 0 1 0 0 0 0 1 1 01 1 1 0 0 0 0 0 0 0

which has determinant 48 6= 0. Therefore, the derivative of lengths d` hasfull rank at Xθ whenever θ 6= π/2. Note that this only gives us that Xg hascodimension at most 19 in Tg, hence dimension at least 11 = 4g − 13. Theconclusion is weaker than that of Theorem 7.1, but we wanted to include

this example to show that D can have full rank for more complicated graphs.

Questions. We conclude with a few questions related to the strategy wehave just outlined.

18 MAXIME FORTIER BOURQUE

Question 7.2. Is there a sequence of graphs M and G as in Section 3 with

girth going to infinity such that the corresponding intersection matrices Dhave non-zero determinants?

In view of the above reasoning and the counting of Proposition 4.1, a po-sitive answer would imply Conjecture 1.2. A major difficulty is that M andG are given to us in a non-explicit way from Theorem 2.1 and Theorem 3.2.

As the proof of Theorem 7.1 shows, one could bypass the determinantissue by finding a filling subset R ⊂ S of even cardinality such that thecorresponding intersection graph IR is a tree, and a surface Y near Xθ

whose systoles are exactly the curves in R.

Question 7.3. Given a surface X constructed as in Section 3 with set ofsystoles S, is there an induced subtree in IS with an even number of verticessuch that the union of the corresponding curves fill?

Question 7.4. Let X be any hyperbolic surface, let S be its set of systolesand let R ⊂ S be a non-empty subset. Does there exist, in every neighbor-hood of X, a surface whose set of systoles is equal to R?

Even if these questions have negative answers, they suggest how oneshould modify the construction of surfaces with sublinearly many systolesthat fill in order to show that Xg has large dimension: the systoles shouldcut the surface into a single polygon instead of several.

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School of Mathematics and Statistics, University of Glasgow, UniversityPlace, Glasgow, United Kingdom, G12 8QQ

E-mail address: maxime.fortier-bourque@glasgow.ac.uk